Block on a Rough Incline with a Hanging Mass
A block of mass 4.0 kg rests on a rough plane inclined at 25° to the horizontal. The coefficient of friction between the block and the plane is 0.30. A light inextensible string attached to the block runs parallel to the plane, passes over a smooth pulley at the top of the plane, and carries a hanging mass of m kg. The system is released from rest, but in the first part it is held in equilibrium. Take g = 9.81 m/s² and ignore air resistance. The pulley is smooth and the string does not slip. You may assume the block remains on the plane and that the plane is long enough for any motion considered. A student claims that every hanging mass between 0.5 kg and 3.0 kg will keep the block stationary. Your task is to test that claim with limiting-equilibrium reasoning, not to assume it is true.
1. Determine the range of values of m for which the block can remain in equilibrium. Then state whether the student’s claim is correct.
Read the worked answer
Take the positive direction as up the plane. The component of the block’s weight down the plane is 4g sin 25° = 4 × 9.81 × 0.4226 = 16.58 N. The normal reaction is 4g cos 25° = 4 × 9.81 × 0.9063 = 35.57 N, so the limiting friction is 0.30 × 35.57 = 10.67 N.
At the lower limit the block is about to slide down the plane, so friction acts up the plane and the tension is just enough to hold it: T_min + 10.67 = 16.58, so T_min = 5.91 N. For the hanging mass in equilibrium, T = mg, so m_min = 5.91 / 9.81 = 0.602 kg.
At the upper limit the block is about to slide up the plane, so friction acts down the plane: T_max = 16.58 + 10.67 = 27.25 N, so m_max = 27.25 / 9.81 = 2.78 kg.
Equilibrium is therefore possible for 0.602 kg ≤ m ≤ 2.78 kg. The student’s claim that every value from 0.5 kg to 3.0 kg works is false: 0.5 kg and 3.0 kg both lie outside the limiting range. Between the limits, friction takes whatever value is needed and is not necessarily at its maximum.
2. The hanging mass is now set to 3.5 kg. Find the acceleration of the system and the tension in the string, stating the direction of the friction force.
Read the worked answer
With m = 3.5 kg, the hanging weight is 3.5 × 9.81 = 34.34 N. This exceeds the maximum tension of 27.25 N that the block can sustain without sliding up, so the block accelerates up the plane and friction acts down the plane at its limiting value.
Let a be the acceleration up the plane and T the tension. For the block: T − 16.58 − 10.67 = 4.0a. For the hanging mass, taking downward as positive: 3.5g − T = 3.5a.
Adding the equations eliminates T: 34.34 − 27.25 = 7.5a, so a = 7.09 / 7.5 = 0.945 m/s².
Then T = 34.34 − 3.5 × 0.945 = 34.34 − 3.31 = 31.03 N. A check using the block equation gives T = 27.25 + 4.0 × 0.945 = 31.03 N, which is consistent. The tension is greater than the equilibrium maximum because the hanging mass is accelerating downward.